I have two sets of known points in $\mathbb{R}^2$: Four points $\mathbf{p_1}, \mathbf{p_2}, \mathbf{p_3}, \mathbf{p_x}$ , and three other points $\mathbf{q_1}, \mathbf{q_2}, \mathbf{q_3}$. I would like to find $\mathbf{q_x}$ such that the ratios of Euclidian distances between $\mathbf{q_x}$ and $\mathbf{q_1}, \mathbf{q_2}, \mathbf{q_3}$, are the same as between $\mathbf{p_x}$ and $\mathbf{p_1}, \mathbf{p_2}, \mathbf{p_3}$. That is, I want to find $\mathbf{q_x}$ such that
$\frac{d(\mathbf{q_1},\mathbf{q_x})}{\sum_i d(\mathbf{q_i},\mathbf{q_x})} = \frac{d(\mathbf{p_1},\mathbf{p_x})}{\sum_i d(\mathbf{p_i},\mathbf{p_x})}$ and $\frac{d(\mathbf{q_2},\mathbf{q_x})}{\sum_i d(\mathbf{q_i},\mathbf{q_x})} = \frac{d(\mathbf{p_2},\mathbf{p_x})}{\sum_i d(\mathbf{p_i},\mathbf{p_x})}$ and $\frac{d(\mathbf{q_3},\mathbf{q_x})}{\sum_i d(\mathbf{q_i},\mathbf{q_x})} = \frac{d(\mathbf{p_3},\mathbf{p_x})}{\sum_i d(\mathbf{p_i},\mathbf{p_x})},$
where $d(\cdot,\cdot)$ is the Euclidian distance. Or to rewrite it a bit nicer:
$d(\mathbf{q_x},\mathbf{q_k}) = w \cdot d(\mathbf{p_x},\mathbf{q_k}) $ for $ k = 1,2,3$,
where $w$ is the unknown factor $w = \frac{\sum_i d(\mathbf{q_i},\mathbf{q_x})}{\sum_i d(\mathbf{p_i},\mathbf{p_x})}$. Note also the more to the point problem formulation by Blue in the comments.
This looks like three equations with three unknowns (x-, and y- coordinates of $\mathbf{q_x}$ and $w$). However, I have been struggling for quite a while now to find out if there is always a unique solution, and to find this solution, or to find an optimal solution if no solution exists.
Any help that you can offer is greatly appreciated!
I'll work from the re-formulation I gave in a comment, but with a minor notational changes.
Given non-negative $u$, $v$, $w$ such that $u+v+w=1$, and given a non-degenerate $\triangle ABC$, we seek a point $P$ such that $$\frac{|PA|}{u} = \frac{|PB|}{v} = \frac{|PC|}{w} = |PA|+|PB|+|PC| =: r \qquad (1)$$
For convenience, introduce these coordinates: $$A = (0,0) \qquad B = (c, 0) \qquad C = ( b \cos A, b \sin A ) \qquad P=(x,y)$$
where $b$, $c$ (and, below, $a$) are lengths of the sides opposite angles $B$, $C$ (and $A$) in $\triangle ABC$. Then, from $(1)$, write
$$\begin{align} (x-0)^2 + ( y - 0 )^2 = r^2 u^2 \quad &\to \quad x^2 + y^2 = r^2 u^2 \\ (x-c)^2 + ( y - 0 )^2 = r^2 v^2 \quad &\to \quad x^2+y^2-2xc + c^2 = r^2 v^2 \\ (x-b \cos A)^2 + ( y - b \sin A )^2 = r^2 w^2 \quad &\to \quad x^2+y^2-2xb\cos A - 2 y b \sin A+ b^2 = r^2 w^2 \\ \end{align}$$
We can use the first equation to eliminate $x^2+y^2$ from the other two equations. (In effect, we're determining the radical axis of each pair of circles.) The result is a linear system $$\begin{align} 2 x c &= c^2 + r^2 \left( u^2 - v^2 \right) \\ 2 x b \cos A + 2 y b \sin A &= b^2 + r^2 \left( u^2 - w^2 \right) \end{align}$$
whose solution is (after some manipulation that probably needs to be double-checked) $$\begin{align} x &= \frac{1}{2c}\left( c^2 + r^2 \left( u^2 - v^2 \right) \right) \\ y &= \frac{1}{2bc\sin A}\left( a b c \cos C + r^2 \left( u^2 a \cos B + v^2 b \cos A - w^2 c \right) \right) \end{align}$$
Substituting these coordinates into the first circle equation gives a quadratic equation in $r^2$:
$$\begin{align} 0 &= r^4 \; \left( \; a^2 ( v^2 - u^2 ) ( u^2 - w^2 ) + b^2 ( w^2 - v^2 ) ( v^2 - u^2 ) + c^2 ( u^2 - w^2 ) ( w^2 - v^2 ) \; \right) \\ &+ 2 r^2 a b c \; \left( \; u^2 a \cos A + v^2 b \cos B + w^2 c \cos C \; \right) \\ &- a^2 b^2 c^2 \end{align}$$
If the coefficient of $r^4$ vanishes ---for instance, when (and only when?) $u=v=w$--- we have $$r^2 = \frac{a b c}{2\; \left( \; u^2 a \cos A + v^2 b \cos B + w^2 c \cos C \; \right)} \qquad (2) $$
Otherwise, we invoke the Quadratic Formula, noting that the discriminant factors nicely as $$\begin{align} &\phantom{\cdot} \left(a+b+c\right)\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right) \\ &\cdot\left(ua+vb+wc\right)\left(-ua+vb+wc\right)\left(ua-vb+wc\right)\left(ua+vb-wc\right) \end{align}$$
The first set of factors constitute Heron's Formula for $16T^2$, where $T$ is the area of $\triangle ABC$. The second set of factors can be interpreted as $16{T_\star}^2$, where $T_\star$ is the area of an auxiliary triangle whose sides are $ua$, $vb$, $wc$ (if such a triangle exists!). We conclude that $$r^2 = \frac{\pm 8 T T_\star - a b c \; \left( \; u^2 a \cos A + v^2 b \cos B + w^2 c \cos C \; \right)}{a^2 ( v^2 - u^2 ) ( u^2 - w^2 ) + b^2 ( w^2 - v^2 ) ( v^2 - u^2 ) + c^2 ( u^2 - w^2 ) ( w^2 - v^2 )} \qquad (3)$$
With an appropriate choice for "$\pm$", we can substitute-back into the formulas for $x$ and $y$ to find $P$. That messy business is left as an exercise for the reader.
I'm not prepared to assert that $u=v=w$ is the only circumstance under which the coefficient of $r^4$ vanishes, putting equation $(2)$ into play. Nevertheless, note that when the coefficient doesn't vanish, solution to the problem requires that $$\left(ua+vb+wc\right)\left(-ua+vb+wc\right)\left(ua-vb+wc\right)\left(ua+vb-wc\right) \ge 0$$
(so that $r^2$ is real). This equation is equivalent the the existence of the auxiliary triangle with sides $ua$, $vb$, $wc$, which is therefore necessary for the existence of $P$. Confirmation (or refutation) that it is also sufficient is another exercise for the reader.
As for uniqueness of the solution (when it exists): The independence of the linear system would seem to guarantee that.